Question

solve this differential equation:

y'+(4/x)y = 2

initial condition: y(1)=-4

Answer 1

this is linear
do you know how to use an integrating factor?

Answer 2

yes, i have solved it but it has been a long time and I can't remember how to handle the constant c that I get out of my integral

Answer 3

how far did you get?

Answer 4

I have an answer, but when I get to this part:

yx^4=(2/5)x^5+c

I'm not sure if I also apply my x^-4 to the c or leave it as a constant

Answer 5

also, how do I handle the absolute value that gets kicked out of a logarithmic function?

Answer 6

if you want the "explicit" solution, solve for y
y(1)=-4 means you can plug in x=1 and y=-4, which will allow you to solve for c

Answer 7

yes, but am I solving y=(2/5)x+c or am i solving y=(2/5)x+cx^-4?

Answer 8

y=(2/5)x+cx^-4?

Answer 9

from this: yx^4=(2/5)x^5+c

do I go to y=(2/5)x + c

or y=(2/5)x+cx^-4?

Answer 10

which can also be written as y=(2/5)x+c/(x^4)

Answer 11

yes, that is correct

Answer 12

which one? the second one?

Answer 13

oh I see, I didn't mean to copy the '?'

the second one is right

y=(2/5)x+cx^-4=(2/5)x+c/(x^4)

Answer 14

gotcha, thank you! I appreciate that. makes life a whole lot easier. lol

Answer 15

from there applying initial conditions is a piece of cake!

Answer 16

thanks for your help man(or woman), I really appreciate it.

Answer 17

no problem